Halves

$Let\quad \theta =\cfrac { \pi }{ 3 } \quad ,\quad then\quad \cos { \theta } =\cfrac { 1 }{ 2 } \\ \sqrt { \cfrac { 1 }{ 2 } +\cfrac { 1 }{ 2 } (\cfrac { 1 }{ 2 } ) } =\sqrt { \cfrac { 1 }{ 2 } +\cfrac { 1 }{ 2 } (\cos { \theta } ) } =\cos { \cfrac { \theta }{ 2 } } \\ Hence\quad the\quad expression\quad becomes:\\ \lim _{ n\rightarrow \infty }{ \cos { \theta } \cos { \cfrac { \theta }{ 2 } } \cos { \cfrac { \theta }{ 4 } ...\cos { \cfrac { \theta }{ { 2 }^{ n } } } } } =\lim _{ n\rightarrow \infty }{ \cos { \theta } \cos { \cfrac { \theta }{ 2 } } \cos { \cfrac { \theta }{ 4 } ...\cos { \cfrac { \theta }{ { 2 }^{ n } } } } } \cdot \cfrac { \sin { \cfrac { \theta }{ { 2 }^{ n } } } }{ \sin { \cfrac { \theta }{ { 2 }^{ n } } } } \\ Since\quad \sin { \cfrac { \theta }{ { 2 }^{ n } } } \cos { \cfrac { \theta }{ { 2 }^{ n } } } =\cfrac { 1 }{ 2 } \sin { \cfrac { \theta }{ { 2 }^{ n-1 } } } ,\quad the\quad expression\quad is\quad now:\\ \lim _{ n\rightarrow \infty }{ \cfrac { \sin { 2\theta } }{ { 2 }^{ n+1 }\sin { \cfrac { \theta }{ { 2 }^{ n } } } } } =\lim _{ n\rightarrow \infty }{ \cfrac { \sin { 2\theta } }{ { 2 }^{ n+1 }\sin { \cfrac { \theta }{ { 2 }^{ n } } } } } (\cfrac { \theta }{ { 2 }^{ n } } )(\cfrac { { 2 }^{ n } }{ \theta } )=\lim _{ n\rightarrow \infty }{ \cfrac { \cfrac { \theta }{ { 2 }^{ n } } }{ \sin { \cfrac { \theta }{ { 2 }^{ n } } } } } \cdot \cfrac { \sin { 2\theta } }{ { 2 }\theta } =\cfrac { \sin { 2\theta } }{ { 2 }\theta } =\cfrac { 3\sqrt { 3 } }{ 4\pi } \\ \therefore \boxed { 3+3+4=10 }$

Similar solution with @Rui-Xian Siew's

Let the infinite product be $P$ . Considering the identity: $\cos \theta = 2 \cos^2 \frac \theta 2 - 1$ $\implies \cos \frac \theta 2 = \sqrt{\frac 12 + \frac 12 \cos \theta}$ . Then, we have:

$\begin{aligned} P & = \frac 12 \cdot \sqrt{\frac 12 + \frac 12 \cdot \frac 12} \cdot \sqrt{\frac 12 + \frac 12 \sqrt{\frac 12 + \frac 12 \cdot \frac 12}} \cdots & \small \color{#3D99F6} \text{Since }\cos \frac \pi 3 = \frac 12 \\ & = \cos \frac \pi 3 \cdot \cos \frac \pi{2\cdot 3} \cdot \cos \frac \pi{2^2 \cdot 3} \cdots \\ & = \lim_{n \to \infty} \prod_{k=0}^n \cos \frac \pi{2^k \cdot 3} \\ & = \lim_{n \to \infty} \frac {\sin \frac \pi{2^n \cdot 3} \cos \frac \pi{2^n \cdot 3}}{\sin \frac \pi{2^n \cdot 3}}\prod_{k=0}^{n-1} \cos \frac \pi{2^k \cdot 3} \\ & = \lim_{n \to \infty} \frac {\sin \frac \pi{2^{n-1} \cdot 3} \cos \frac \pi{2^{n-1} \cdot 3}}{2 \sin \frac \pi{2^n \cdot 3}}\prod_{k=0}^{n-2} \cos \frac \pi{2^k \cdot 3} \\ & = \lim_{n \to \infty} \frac {\sin \frac \pi{2^{n-2} \cdot 3} \cos \frac \pi{2^{n-2} \cdot 3}}{2^2 \sin \frac \pi{2^n \cdot 3}}\prod_{k=0}^{n-3} \cos \frac \pi{2^k \cdot 3} \\ & = \ \cdots \\ & = \lim_{n \to \infty} \frac {\sin \frac {2\pi}3}{2^{n+1} \sin \frac \pi{2^n\cdot 3}} \\ & = \lim_{n \to \infty} \frac 12 \cdot \sin \frac {2\pi}3 \cdot \left(\frac \pi 3 \cdot\color{#3D99F6} \frac {\sin \frac \pi{2^n\cdot 3}}{\frac \pi{2^n \cdot 3}}\right)^{-1} & \small \color{#3D99F6} \text{Note that } \lim_{x\to 0} \frac {\sin x}x = 1 \\ & = \frac {3\sqrt 3}{4\pi} \end{aligned}$

Therefore, $a+b+c = 3+3+4 = \boxed {10}$ .